$A$ data consists of $n$ observations $x_1, x_2, ......, x_n$. If $\sum_{i=1}^n (x_i + 1)^2 = 9n$ and $\sum_{i=1}^n (x_i - 1)^2 = 5n$,then the standard deviation of this data is

Let $a_1, a_2, \ldots, a_{10}$ be $10$ observations such that $\sum_{k=1}^{10} a_k = 50$ and $\sum_{k < j} a_k a_j = 1100$. Then the standard deviation of $a_1, a_2, \ldots, a_{10}$ is equal to:

The sum and sum of squares corresponding to length $x$ (in $cm$) and weight $y$ (in $gm$) of $50$ plant products are given below:
$\sum\limits_{i = 1}^{50} {{x_i} = 212, \sum\limits_{i = 1}^{50} {x_i^2} = 902.8, \sum\limits_{i = 1}^{50} {{y_i} = 261, \sum\limits_{i = 1}^{50} {y_i^2 = 1457.6} } }$
Which is more varying,the length or weight?

Let population $A$ contain $100$ observations $101, 102, \dots, 200$ and another population $B$ contain $100$ observations $151, 152, \dots, 250$. If $V_A$ and $V_B$ represent the variances of the two populations respectively,then what is $V_A / V_B$?

If the variance of the first $n$ natural numbers is $10$ and the variance of the first $m$ even natural numbers is $16$,then $n: m=$

The mean and variance of $10$ observations were calculated as $15$ and $15$ respectively by a student who took by mistake $25$ instead of $15$ for one observation. Then,the correct standard deviation is $.....$